Chemistry: Principles and Reactions
Chemistry: Principles and Reactions
8th Edition
ISBN: 9781305079373
Author: William L. Masterton, Cecile N. Hurley
Publisher: Cengage Learning
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Chapter 15, Problem 34QAP

Solid lead nitrate is added to a solution that is 0.020 M in OH- and SO42-. Addition of the lead nitrate does not change the volume of the solution.

(a) Which compound, PbSO4 or Pb(OH)2 ( K sp = 2.8 × 10 16 ) , will precipitate first?

(b) What is the pH of the solution when PbSO4 first starts to precipitate?

(a)

Expert Solution
Check Mark
Interpretation Introduction

Interpretation: Whether PbSO4orPb(OH)2(Ksp=2.8×10-16), will precipitate first needs to be determined.

Concept Introduction :

PbSO4 (Equilibrium reaction)

PbSO4(s)Pb+2(aq)+SO42(aq)

Expression of Pb+2 required for precipitation of PbSO4 is −

Ksp=[Pb+2][SO42]...... (1)

Where, Ksp is solubility product of PbSO4 , [Pb+2] is concentration of lead ion and [SO42] concentration of sulfate ion.

  Pb(OH)2 (Equilibrium reaction)

  Pb(OH)2(s)Pb+2(aq)+2OH(aq)

Expression of Pb+2 required for precipitation of Pb(OH)2 is −

Ksp=[Pb+2][OH]2...... (2)

Where, Ksp is solubility product of Pb(OH)2 , [Pb+2] is concentration of lead ion and [OH] concentration of hydroxide ion.

Answer to Problem 34QAP

The compound which precipitates first is Pb(OH)2 .

Explanation of Solution

Given Information:

OH=0.020MSO42=0.020M

The value of solubility product for both the salts is 2.8×1016 .

Calculation:

  PbSO4 (Equilibrium reaction) is as follows:

PbSO4(s)Pb+2(aq)+SO42(aq)

Expression of Pb+2 required for precipitation of PbSO4 is −

Ksp=[Pb+2][SO42]...... (1)

Where, Ksp is solubility product of PbSO4 , [Pb+2] is concentration of lead ion and [SO42] concentration of sulfate ion.

  Ksp(PbSO4)=2.8×1016

SO42=0.020M

  [Pb+2] is find out by putting above values in equation (1).

  Ksp=[Pb+2][SO42]2.8×1016=[Pb+2]×0.020[Pb+2]=2.8× 10 160.020[Pb+2]=1.4×1014M

So, [Pb+2] concentration is 1.4×1014M .

  Pb(OH)2 (Equilibrium reaction) is as follows:

Pb(OH)2(s)Pb+2(aq)+2OH(aq)

Expression of

Pb+2 required for precipitation of Pb(OH)2 is −

Ksp=[Pb+2][OH]2...... (2)

Where, Ksp is solubility product of Pb(OH)2 , [Pb+2] is concentration of lead ion and [OH] concentration of hydroxide ion.

  Ksp(Pb( OH)2)=2.8×1016

[OH]=0.020M

  [Pb+2] is find out by substitution of above values in equation (2).

  Ksp=[Pb+2][ OH]22.8×1016=[Pb+2]×(0.020)2[Pb+2]=2.8× 10 160.0004[Pb+2]=7×1013M

So, [Pb+2] concentration is 7×1013M .

[Pb+2] Required for precipitation of PbSO4 is 1.4×1014M which is less than [Pb+2] required for precipitation of Pb(OH)2 that is 7×1013M .

Hence, Pb(OH)2 precipitates first.

(b)

Expert Solution
Check Mark
Interpretation Introduction

Interpretation:The pH of the solution when PbSO4 first starts to precipitate needs to be determined.

Concept Introduction :

Formula to calculate pH from pOH value is as follows:

pH=14pOH

Formula to calculate pOH is as follows:

pOH=log[OH]

Answer to Problem 34QAP

When PbSO4 first starts to precipitate pH of the solution is 9.24 .

Explanation of Solution

When PbSO4 first start to precipitate, value of [OH] can be calculated as follows:

From Pb(OH)2 dissociation reaction,

  Ksp=[Pb+2][ OH]2[OH]=( K sp [ Pb +2 ])12[OH]=( 2.8× 10 16 9× 10 7 )12=0.0000176M

So, [OH] is 0.0000176M

Calculation of pOH .

pOH=log[OH]=log(0.0000176)=4.75

So, value of pOH is 4.75 .

Calculation of pOH of solution as follows:

  pH=14pOH=144.75=9.24

Hence, pH of the solution is 9.24 .

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Chapter 15 Solutions

Chemistry: Principles and Reactions

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